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MD
giúp mình với mình chỉ cần 2 câu 8 với câu 9 thôi nha
2
TP
Thảo Phương
CTVVIP
16 tháng 8 2021
Câu 8 :
a) \(n_{Cu}=\dfrac{6,4}{64}=0,1\left(mol\right)\)
\(n_{Al}=\dfrac{54}{27}=2\left(mol\right)\)
b) \(V_{CO_2}=0,15.22,4=3,36\left(l\right)\)
\(V_{N_2}=0,3.22,4=6,72\left(l\right)\)
c) \(n_{hh}=n_{CO_2}+n_{H_2}=\dfrac{0,22}{44}+\dfrac{0,02}{2}=0,015\left(mol\right)\)
\(V_{hh}=0,015.22,4=0,336\left(l\right)\)
TP
Thảo Phương
CTVVIP
16 tháng 8 2021
Câu 9
a) \(m_N=0,3.14=4.2\left(g\right)\)
\(m_{Cl}=0,4.35,5=14,2\left(g\right)\)
\(m_O=5.16=80\left(g\right)\)
b) \(m_{N_2}=0,2.28=5,6\left(h\right)\)
\(m_{Cl_2}=0,3.71=21,3\left(g\right)\)
\(m_{O_2}=4.32=128\left(g\right)\)
c) \(m_{Fe}=0,12.56=6,72\left(g\right)\)
\(m_{Cu}=3,15.64=201,6\left(g\right)\)
\(m_{H_2SO_4}=0,85.98=83,3\left(g\right)\)
\(m_{CuSO_4}=0,52.160=83,2\left(g\right)\)
VL
18 tháng 4 2019
Science and technology has become a debated topic in the society. On one hand, it is necessary for the modern life where other countries are continuously developing in the field of science and technology. It becomes very necessary for other countries too to grow in the same way to be strong and well developed other countries for the future safety and security. It is science and technology which helps other weak countries to develop and be strong. We have to take support of science and technology forever to improve the way of life for the betterment of mankind. If we do not take the help of technologies such as computer, internet, electricity, etc we cannot be economically strong in the future and would be backward forever even we cannot survive in such a competitive and technological world.
Advancement in the field of medical, agriculture, education, economy, sports, games, jobs, tourism, etc are the examples of science and technology. All such advancements show us that how both are equally beneficial for our life. We can see a clear difference in our life style while matching the ancient and modern way of life. High level of scientific and technological advancement in the field of medicine has made easy the treatment of various lethal diseases which was earlier not possible. It has helped a lot to the doctors to find effective ways to cure diseases through medicine or operations as well as research vaccines to cure diseases such as cancer, AIDS, diabetes, Alzheimer’s, Leukemia, etc.
17 tháng 3 2022
nNaOH = 16/40 = 0,4 (mol)
PTHH: 2NaOH + SO2 -> Na2SO3 + H2O
Mol: 0,4 ---> 0,2 ---> 0,2
VSO2 = 0,2 . 22,4 = 4,48 (l)
mNa2SO3 = 0,2 . 126 = 25,2 (g)
T
3
giúp mik với, mình có mỗi 30p thôi
14 tháng 1 2022
$a)PTHH:2Al+6HCl\to 2AlCl_3+3H_2$
$n_{Al}=\dfrac{5,4}{27}=0,2(mol)$
$\Rightarrow n_{H_2}=1,5.n_{Al}=0,3(mol)$
$\Rightarrow V_{H_2}=0,3.22,4=6,72(l)$
$b)n_{AlCl_3}=n_{Al}=0,2(mol)$
$\Rightarrow m_{AlCl_3}=0,2.133,5=26,7(g)$
$c)n_{HCl}=3n_{Al}=0,6(mol)$
$\Rightarrow m_{HCl}=0,6.36,5=21,9(g)$
DT
14 tháng 1 2022
pthh :
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
a,
\(nAl=\dfrac{5,4}{27}=0,2mol\)
tính theo pthh: \(nH_2=\dfrac{3}{2}.nAl=\dfrac{3}{2}.0,2=0,3mol\)
\(VH_2=0,3.22,4=6,72lít\)
b,
tính theo pthh :
\(nAlCl_3=\dfrac{2}{2}.nAl=1.0,2=0,2mol\\ \Rightarrow mAlCl_3=0,2.133,5=26,7gam\)
c,
tính theo pthh :
\(nHCl=\dfrac{6}{2}.nAl=3.0,2=0,6mol\)
\(\Rightarrow mHCl=0,6.36,5=21,9gam\)
NL
0
KM
1
H9
HT.Phong (9A5)
CTVHS
22 tháng 3 2023
BT4: Hiệu suất phản ứng:
\(H=\dfrac{m_{tt}}{m_{lt}}.100\%=\dfrac{36,48}{48}.100\%=76\%\)
BT5 Khối lượng đồng thu được:
\(H=\dfrac{m_{tt}}{m_{lt}}.100\%\Rightarrow m_{tt}=\dfrac{m_{lt}.H}{100\%}=\dfrac{48.95}{100\%}=45,6\left(g\right)\)
13 tháng 7 2016
Đặt số hạt proton, nơtron và electron trong nguyên tử X lần lượt là p,n,e (p,n,e \(\in N\) sao)
Theo ĐB ta có: p+n+e=52
p+e-n=16
\(\Rightarrow\begin{cases}2p+n=52\\2p-n=16\end{cases}\)
\(\Rightarrow\begin{cases}p=17\Rightarrow e=17\\n=18\end{cases}\)
Bài 3:
Ta có: \(M_{X_2O_5}=\dfrac{21,3}{0,15}=142\left(g/mol\right)\)
⇒ \(M_X=\left(M_{X_2O_5}-5M_O\right):2=31\left(g/mol\right)\)
⇒ X là nguyên tố photpho (P)
Vậy CTHH:P2O5
Bài 4:
Ta có: \(M_{Y_2O_5}=\dfrac{5,4}{0,05}=108\left(g/mol\right)\)
⇒ \(M_Y=\left(108-5.16\right):2=14\left(g/mol\right)\)
⇒ Y là nguyên tố nitơ (N)
Vậy CTHH:N2O5